Only $$–CHO$$ group reacts with \[C{{N}^{-}}\,ion\] and the reaction is
43.
Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using $$NaOH.$$ The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :
$$CC{l_3}CHO + NaOH \to \mathop {CC{l_3}C{H_2}OH + CC{l_3}COONa}\limits_{{\text{2, 2,2 }} - \,\,{\text{trichloroethanol}}} $$
In Cannizzaro’s reaction the compounds which do not contain $$\alpha $$ - hydrogen atoms undergo oxidation and reduction simultaneously i.e undergo disproportion ation and form one molecule of sodium salt of carboxylic acid as oxidation product and one molecule of alcohol as reduction product.
44.
Consider the reactions,
Identify $$A, X,Y$$ and $$Z$$
Aldehydes gives silver minor test so, $$'X’$$ may be alcohol which is oxidised by $$Cu$$ gives aldehydes.
Therefore,
A is acetaldehyde $$\left( {C{H_3}CHO} \right)$$
45.
Hydrocarbons are formed when aldehydes and ketones are reacted with amalgamated zinc and conc. $$HCl.$$ The reaction is called
When benzaldehyde is treated with 50% alkali, it undergoes oxidation to give an acid salt as well as reduction to give an alcohol. This reaction is called Cannizaro’s reaction
47.
Fill in the blanks by suitable choices.
The carbon atom in carbonyl group is ______ hybridised. The carbonyl group in aldehydes and ketones undergoes ______ reactions. Aldehydes which have $$\alpha {\text{ - }}H$$ atom undergo ______ reaction while aldehydes which have no $$\alpha {\text{ - }}H$$ atom undergo ______ reaction.
A
$$s{p^3},$$ nucleophilic substitutions, aldol condensation, Cannizzaro
B
$$sp,$$ electrophilic substitution, Cannizzaro, aldol condensation
C
$$s{p^2},$$ nucleophilic addition, aldol condensation, Cannizzaro
D
$$s{p^3},$$ electrophilic addition, Cannizzaro, aldol condensation
Unsaturation is not detected by Tollen’s reagent, a reagent for differentiating between aldehydes and ketones.
49.
A substance $${C_4}{H_{10}}O$$ yields on oxidation a compound, $${C_4}{H_8}O$$ which gives an oxime and a positive iodoform test. The original substance on treatment with $$conc.\,{H_2}S{O_4}$$ gives $${C_4}{H_8}.$$ The structure of the compound is
\[{{C}_{4}}{{H}_{8}}\xleftarrow[\left( -{{H}_{2}}O \right)]{Conc.\,{{H}_{2}}S{{O}_{4}}}{{C}_{4}}{{H}_{10}}O\xrightarrow{\text{Oxidation}}{{C}_{4}}{{H}_{8}}O\left( R-COC{{H}_{3}} \right)\]
Thus \[{{C}_{4}}{{H}_{8}}O\] should be \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{3}},\] hence \[{{C}_{4}}{{H}_{10}}O\] should be \[C{{H}_{3}}C{{H}_{2}}CHOHC{{H}_{3}}\]
50.
Few simple chemical tests are given below to differentiate between the pairs of compounds. Which of the following tests is not correct for differentiation?
A
Propanal and propanone - Silver mirror test
B
Acetophenone and benzophenone - Iodoform test
C
Ethanal and propanal - Fehling's test
D
Benzoic acid and ethyl benzoate - Sodium bicarbonate test